3 comments:

Draw lines per attached sketchWe have ∆CA’B’ and CA”’B” are isosceles => A’A””=B’B”From B draw line C1A1//ACThis line cut B’A’, B’C’,GC”’ and GA”’ at A1, C1, C2 and A2 .Note that ∆CA’B’, ∆ BA’A1, ∆AB’C’ and ∆BC’C1 are isosceles So BC1=BC’=BA’=BA1 => BA1=BC1Since GA2//B’A1 and GC2//B’C1 we have BA2/BA1=BG/BB’=BC2/BC1B is the midpoint of A2C2 => B’ is the midpoint of B”B”’ Triangle IB”B”’ is isosceles Similarly IC”C”’ and IA”A”’ are isosceles => IB”=IB”’=IC”=IC”’=IA”=IA”’So I is the center of circle A”A”’B”B”’C”C

Denote points P1,P2,P3,...P6 as the points where the parallel lines cut the triangles (starting from the right hand side of A)By the properties of incircle, AC'=AB' and since P2P5 parrllel to B'C', AP2 = AP5 hence B'P5 = C'P2.